Solution: Pendulum Confined Between Walls
The natural amplitude of the pendulum is \(\alpha\), and the walls are at angle \(\beta\) where \(\beta < \alpha\). Normally, the pendulum hits the walls and rebounds.
If a collision with even one wall is inelastic (coefficient of restitution \(e < 1\)), energy is dissipated during that impact. This causes the amplitude of oscillation to decrease with every cycle.
As the amplitude reduces, eventually the new amplitude \(\alpha’\) will become less than the wall angle \(\beta\) (\(\alpha’ < \beta\)).
Once \(\alpha’ < \beta\), the pendulum no longer touches the walls. It oscillates freely between angles \(-\alpha’\) and \(+\alpha’\).
The period of a free simple pendulum is independent of amplitude (for small angles). Thus, the period settles to the natural period:
This explains why both (b) and (c) are correct: in both cases, energy loss leads to free oscillation.